What is the d/(dt)((x)(-1.25x^2+5x)) ?

The d/(dt)((x)(-1.25x^2+5x)) is -3.75t^2+10t

What is the first d/(dt)((x)(-1.25x^2+5x)) ?

The first d/(dt)((x)(-1.25x^2+5x)) is -3.75t^2+10t

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\frac{d}{d t} ((t) (- 1.25 t^{2} + 5 t))

- 3.75 t^{2} + 10 t

Solution steps

\frac{d}{d t} ((t) (- 1.25 t^{2} + 5 t))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t, g = - 1.25 t^{2} + 5 t

= \frac{d t}{d t} (- 1.25 t^{2} + 5 t) + \frac{d}{d t} (- 1.25 t^{2} + 5 t) t

\frac{d t}{d t} = 1

\frac{d}{d t} (- 1.25 t^{2} + 5 t) = - 2.5 t + 5

= 1 \cdot (- 1.25 t^{2} + 5 t) + (- 2.5 t + 5) t

Simplify

1 \cdot (- 1.25 t^{2} + 5 t) + (- 2.5 t + 5) t : - 3.75 t^{2} + 10 t

= - 3.75 t^{2} + 10 t

\frac{( - a - 5 ) ^{2}}{5 ( a ^{2} + 5 )}

y, x + 2 y = 1

(x + y) + z

(1 - e^{- 2 t}) u \cdot (t)

9 8^{3}

What is the d/(dt)((x)(-1.25x^2+5x)) ?
The d/(dt)((x)(-1.25x^2+5x)) is -3.75t^2+10t
What is the first d/(dt)((x)(-1.25x^2+5x)) ?
The first d/(dt)((x)(-1.25x^2+5x)) is -3.75t^2+10t